Workspace setup:

As we develop more useful models, we’ll begin to practice the art of generating models with multiple estimands. An estimand is a quantity we want to estimate from the data. Our models may not themselves produce the answer to our central question, so we need to know how to calculate these values from the posterior distributions.

This is going to be different from prior regression courses (PSY 612), where our models were often designed to give us precisely what we wanted. For example, consider the regression:

\[ \hat{Y} = b_0 + b_1(D) \] Where \(Y\) is a continuous outcome and \(D\) is a dummy coded variable (0 = control; 1 = treatment).

• What does \(b_0\) represent?
• What does \(b_1\) represent?
• How would you calculate or estimate the means of both groups from this model?

Categories

Forget dummy codes. From here on out, we will incorporate categorical causes into our models by using index variables. An index variable contains integers that correspond to different categories. The numbers have no inherent meaning – rather, they stand as placeholders or shorthand for categories.

data("Howell1")
d <- Howell1
d$sex <- ifelse(d$male == 1, 2, 1) # 1 = female, 2 = male
head(d[, c("male", "sex")])

  male sex
1    1   2
2    0   1
3    0   1
4    1   2
5    0   1
6    1   2

Mathematical model

Let’s write a mathematical model to express height in terms of sex.

\[\begin{align*} h_i &\sim \text{Normal}(\mu_i, \sigma) \\ \mu_i &= \alpha_{SEX[i]} \\ \alpha_j &\sim \text{Normal}(178, 20)\text{ for }j = 1..2 \\ \sigma &\sim \text{Uniform}(0, 50) \end{align*}\]

flist <- alist(
  height ~ dnorm( mu , sigma) ,
  mu <- a[sex] ,
  a[sex] ~ dnorm( 178, 20 ) ,
  sigma ~ dunif(0, 50)
)

Fitting the model using quap()

m1 <- quap(
  flist, data=d
)

precis(m1, depth=2)

           mean        sd      5.5%     94.5%
a[1]  134.90835 1.6068669 132.34026 137.47643
a[2]  142.57862 1.6973997 139.86585 145.29140
sigma  27.30881 0.8279547  25.98558  28.63204

Here, we are given the estimates of the parameters specified in our model: the average height of women (a[1]) and the average height of men (a[2]). But our question is whether these average heights are different. How do we get that?

Calculate the contrast

We can create two plots. One is the posterior distributions of average female and male heights and one is the average difference.

p1 <- post %>% as.data.frame() %>% 
  pivot_longer(starts_with("a")) %>% 
  mutate(sex = ifelse(name == "a.1", "female", "male")) %>% 
  ggplot(aes(x=value, color = sex)) +
  geom_density(linewidth = 2) +
  labs(x = "height(cm)") 

p2 <- post %>% as.data.frame() %>% 
  ggplot(aes(x=diff_fm)) +
  geom_density(linewidth = 2) +
  labs(x = "difference in height(cm)") 

( p1 | p2)

Expected values vs predicted values

A note that the distributions of the mean heights is not the same as the distribution of heights period. For that, we need the posterior predictive distributions.

pred_f  <- rnorm(1e4, mean = post$a[,1], sd = post$sigma )
pred_m  <- rnorm(1e4, mean = post$a[,2], sd = post$sigma )

pred_post = data.frame(pred_f, pred_m) %>%
  mutate(diff = pred_f-pred_m)

# plot distributions
p1 <- pred_post %>% pivot_longer(starts_with("pred")) %>% 
  mutate(sex = ifelse(name == "pred_f", "female", "male")) %>% 
  ggplot(aes(x = value, color = sex)) +
  geom_density(linewidth = 2) +
  labs(x = "height (cm)")

# plot difference
# Compute density first
density_data <- density(pred_post$diff)

# Convert to a tibble for plotting
density_df <- tibble(
  x = density_data$x,
  y = density_data$y,
  fill_group = ifelse(x < 0, "male", "female")  # Define fill condition
)

# Plot with area fill
p2 <- ggplot(density_df, aes(x = x, y = y, fill = fill_group)) +
  geom_area() +  # Adjust transparency if needed
  geom_line(linewidth = 1.2, color = "black") +  # Keep one continuous curve
  labs(x = "Difference in height (F-M)", y = "density") +
  guides(fill = "none")

(p1 | p2)

exercise

In the rethinking package, the dataset milk contains information about the composition of milk across primate species, as well as some other facts about those species. The taxonomic membership of each species is included in the variable clade; there are four categories.

1. Create variable in the dataset to assign an index value to each of the 4 categories.
2. Standardize the milk energy variable (kcal.per.g). ¹
3. Write a mathematical model to express the average milk energy (in standardized kilocalories) in each clade.

solution

data("milk")
str(milk)

'data.frame':   29 obs. of  8 variables:
 $ clade         : Factor w/ 4 levels "Ape","New World Monkey",..: 4 4 4 4 4 2 2 2 2 2 ...
 $ species       : Factor w/ 29 levels "A palliata","Alouatta seniculus",..: 11 8 9 10 16 2 1 6 28 27 ...
 $ kcal.per.g    : num  0.49 0.51 0.46 0.48 0.6 0.47 0.56 0.89 0.91 0.92 ...
 $ perc.fat      : num  16.6 19.3 14.1 14.9 27.3 ...
 $ perc.protein  : num  15.4 16.9 16.9 13.2 19.5 ...
 $ perc.lactose  : num  68 63.8 69 71.9 53.2 ...
 $ mass          : num  1.95 2.09 2.51 1.62 2.19 5.25 5.37 2.51 0.71 0.68 ...
 $ neocortex.perc: num  55.2 NA NA NA NA ...

milk$clade_id <- as.integer(milk$clade)
milk$K <- standardize(milk$kcal.per.g)

\[\begin{align*} K_i &\sim \text{Normal}(\mu_i, \sigma) \\ \mu_i &= \alpha_{\text{CLADE}[i]} \\ \alpha_i &\sim \text{Normal}(0, 0.5) \text{ for }j=1..4 \\ \sigma &\sim \text{Exponential}(1) \\ \end{align*}\]

Exercise: Now fit your model using quap(). It’s ok if your mathematical model is a bit different from mine.

solution

flist <- alist(
  K ~ dnorm( mu , sigma ) ,
  mu <- a[clade_id] , 
  a[clade_id] ~ dnorm( 0 , 0.5 ) , 
  sigma ~ dexp( 1 )
)

m2 <- quap(
  flist, data = milk
)

precis( m2, depth=2 )

            mean         sd        5.5%      94.5%
a[1]  -0.4843464 0.21764156 -0.83217966 -0.1365132
a[2]   0.3662511 0.21705924  0.01934847  0.7131537
a[3]   0.6752371 0.25753382  0.26364833  1.0868259
a[4]  -0.5858007 0.27450960 -1.02452009 -0.1470814
sigma  0.7196466 0.09653382  0.56536687  0.8739263

Plotting with rethinking

labels <- paste( "a[" , 1:4, "]:", levels(milk$clade),  sep="" )
plot(
  precis(m2, depth=2, pars = "a"),
  labels=labels, 
  xlab="expected kcal (std)"
)

exercise

Plot the following distributions:

• Posterior distribution of average milk energy by clade.
• Posterior distribution of predicted milk energy values by clade.

solution

post <- extract.samples( m2 )
names(labels) = paste("a.", 1:4, sep = "")
post %>% as.data.frame() %>% 
  pivot_longer(starts_with("a")) %>% 
  mutate(name = recode(name, !!!labels)) %>% 
  ggplot(aes(x = value, color = name)) +
  geom_density(linewidth = 2) +
  labs(title = "Posterior distribution of expected milk energy")

solution

post <- extract.samples( m2 )
a.1 = rnorm(1e4, post$a[,1], post$sigma)
a.2 = rnorm(1e4, post$a[,2], post$sigma)
a.3 = rnorm(1e4, post$a[,3], post$sigma)
a.4 = rnorm(1e4, post$a[,4], post$sigma)
data.frame(a.1, a.2, a.3, a.4) %>% 
  pivot_longer(everything()) %>% 
  mutate(name = recode(name, !!!labels)) %>% 
  ggplot(aes(x = value, color = name)) +
  geom_density(linewidth = 2) +
  labs(title = "Posterior distribution of predicted milk energy")

Combining index variables and slopes

Let’s return to the height example. What if we want to control for weight?

\[\begin{align*} h_i &\sim \text{Normal}(\mu_i, \sigma) \\ \mu_i &= \alpha_{S[i]} + \beta_{S[i]}(W_i-\bar{W})\\ \alpha_j &\sim \text{Normal}(178, 20)\text{ for }j = 1..2 \\ \beta_j &\sim \text{Normal}(0, 5)\text{ for }j = 1..2 \\ \sigma &\sim \text{Uniform}(0, 50) \end{align*}\]

dat <- list(
  height = d$height,
  weight = d$weight,
  Wbar <- mean(d$weight),
  sex = d$male + 1
)

flist <- alist(
  height ~ dnorm( mu , sigma) ,
  mu <- a[sex] + b[sex]*(weight-Wbar),
  a[sex] ~ dnorm( 178, 20 ) ,
  b[sex] ~ dnorm( 0, 20 ) ,
  sigma ~ dunif(0, 50)
)

m3 <- quap(flist, data=dat)

Plot the slopes using extract.samples()

xbar = mean(d$weight)
plot(NULL, xlim = range(d$weight), ylim = c(0, 200),
     xlab = "weight", ylab = "height")
#plot each line
for(i in 1:50){
 curve(post$a[i, 1] +post$b[i, 1]*(x-xbar), 
       add = T,
       col=col.alpha("#1c5253",0.1))  
  curve(post$a[i, 2] +post$b[i, 2]*(x-xbar), 
       add = T,
       col=col.alpha("#e07a5f",0.1))  
}

Plot the slopes using link()

xseq <- seq( min(d$weight), max(d$weight), len=100)
plot(NULL, xlim = range(d$weight), ylim = range(d$height), xlab = "weight", ylab = "height")
muF <- 
  link(m3, data=list(sex=rep(1,100), weight=xseq, Wbar = mean(d$weight)))
lines(xseq, apply(muF, 2, mean), lwd = 2, col = "#1c5253" )
muM <- 
  link(m3, data=list(sex=rep(2,100), weight=xseq, Wbar = mean(d$weight)))
lines(xseq, apply(muM, 2, mean), lwd = 2, col =  "#e07a5f")

exercise

Return to the milk data. Write a mathematical model expressing the energy of milk as a function of the species body mass (mass) and clade category. Be sure to include priors. Fit your model using quap().

solution

\[\begin{align*} K_i &\sim \text{Normal}(\mu_i, \sigma) \\ \mu_i &= \alpha_{\text{CLADE}[i]} + \beta_{\text{CLADE}[i]}(M-\bar{M})\\ \alpha_i &\sim \text{Normal}(0, 0.5) \text{ for }j=1..4 \\ \beta_i &\sim \text{Normal}(0, 0.5) \text{ for }j=1..4 \\ \sigma &\sim \text{Exponential}(1) \\ \end{align*}\]

dat <- list(
  K        = standardize(milk$kcal.per.g),
  M        = milk$mass,
  Mbar     = mean(milk$mass),
  clade_id = milk$clade_id
)

flist <- alist(
  K ~ dnorm( mu , sigma ) ,
  mu <- a[clade_id] +b[clade_id]*(M-Mbar), 
  a[clade_id] ~ dnorm( 0 , 0.5 ) , 
  b[clade_id] ~ dnorm( 0 , 0.5 ) , 
  sigma ~ dexp( 1 )
)

m4 <- quap(
  flist, data = dat
)